使用YQL多查询& XPath解析HTML，如何转义嵌套引号？

Question

标题比它必须的更复杂，这是问题查询。

SELECT * 
FROM query.multi 
WHERE queries="
    SELECT * 
        FROM html 
        WHERE url='http://www.stumbleupon.com/url/http://www.guildwars2.com' 
        AND xpath='//li[@class=\"listLi\"]/div[@class=\"views\"]/a/span';
    SELECT * 
        FROM xml 
        WHERE url='http://services.digg.com/1.0/endpoint?method=story.getAll&link=http://www.guildwars2.com';
    SELECT * 
        FROM json 
        WHERE url='http://api.tweetmeme.com/url_info.json?url=http://www.guildwars2.com';
    SELECT * 
        FROM xml 
        WHERE url='http://api.facebook.com/restserver.php?method=links.getStats&urls=http://www.guildwars2.com';
    SELECT * 
        FROM json 
        WHERE url='http://www.reddit.com/button_info.json?url=http://www.guildwars2.com'"

具体来说，这一行

xpath='//li[@class=\"listLi\"]/div[@class=\"views\"]/a/span'

是有问题的，因为引用，我必须将它们嵌套三层深，并且我已经用完了可以使用的引号字符。我尝试了以下变体但没有成功：

//no attribute quoting
xpath='//li[@class=listLi]/div[@class=views]/a/span' 

//try to quote attribute w/ backslash & single quote
xpath='//li[@class=\'listLi\']/div[@class=\'views\']/a/span'

//try to quote attribute w/ backslash & double quote
xpath='//li[@class=\"listLi\"]/div[@class=\"views\"]/a/span'

//try to quote attribute with double single quotes, like SQL
xpath='//li[@class=''listLi'']/div[@class=''views'']/a/span'

//try to quote attribute with double double quotes, like SQL
xpath='//li[@class=""listLi""]/div[@class=""views""]/a/span'

//try to quote attribute with quote entities
xpath='//li[@class="listLi"]/div[@class="views"]/a/span'

//try to surround XPath with backslash & double quote
xpath=\"//li[@class='listLi']/div[@class='views']/a/span\"

//try to surround XPath with double double quote
xpath=""//li[@class='listLi']/div[@class='views']/a/span""

全部都没有成功。

我没有看到太多关于转义 XPath 字符串的内容，但我发现的所有内容似乎都是使用 concat （这不会有帮助，因为 ' 和 " 都不可用）或 html 实体的变体。不使用属性的引号不会不会抛出错误，但会失败，因为它不是我需要的实际 XPath 字符串，

我在 YQL 文档中没有看到任何有关如何处理转义的内容，但我希望他们会这样做。有某种逃生指南。

Answer 1

您需要使用双反斜杠转义任何分隔 XPath 查询的字符...换句话说：

SELECT * FROM query.multi 
WHERE queries="
    SELECT * 
        FROM html 
        WHERE url='http://www.stumbleupon.com/url/http://www.guildwars2.com' 
        AND xpath='//li[@class=\\'listLi\\']/div[@class=\\'views\\']/a/span';
    SELECT * 
        FROM xml 
        WHERE url='http://services.digg.com/1.0/endpoint?method=story.getAll&link=http://www.guildwars2.com';
    SELECT * 
        FROM json 
        WHERE url='http://api.tweetmeme.com/url_info.json?url=http://www.guildwars2.com';
    SELECT * 
        FROM xml 
        WHERE url='http://api.facebook.com/restserver.php?method=links.getStats&urls=http://www.guildwars2.com';
    SELECT * 
        FROM json 
        WHERE url='http://www.reddit.com/button_info.json?url=http://www.guildwars2.com'"

^{(在 YQL 控制台中尝试此操作）}

Answer 2

我想出了一个解决方案，它并没有真正回答我原来的问题，但确实解决了问题。

data.html.cssselect 表将采用 CSS 选择器和将其解析为 XPath，避免令人讨厌的转义问题。

SELECT *
FROM query.multi 
    WHERE queries="
        SELECT * 
            FROM data.html.cssselect 
            WHERE url='http://www.stumbleupon.com/url/http://www.guildwars2.com' 
            AND css='li.listLi div.views a span';
        SELECT * 
            FROM xml 
            WHERE url='http://services.digg.com/1.0/endpoint?method=story.getAll&link=http://www.guildwars2.com';
        SELECT * 
            FROM json 
            WHERE url='http://api.tweetmeme.com/url_info.json?url=http://www.guildwars2.com';
        SELECT * 
            FROM xml 
            WHERE url='http://api.facebook.com/restserver.php?method=links.getStats&urls=http://www.guildwars2.com';
        SELECT * 
            FROM json 
            WHERE url='http://www.reddit.com/button_info.json?url=http://www.guildwars2.com'"